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Basis Collapse in Holographic Algorithms Over All Domain Sizes - - PowerPoint PPT Presentation

Basis Collapse in Holographic Algorithms Over All Domain Sizes Sitan Chen Harvard College March 29, 2016 Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof


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Basis Collapse in Holographic Algorithms Over All Domain Sizes

Sitan Chen Harvard College March 29, 2016

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Holographic algorithms reduce counting problems into the problem of counting perfect matchings in a graph G = (V, E).

  • Perfect matching: M ⊂ E for which every v ∈ V belongs to

exactly one edge e ∈ M

  • [Valiant ’79]: Counting perfect matchings in arbitrary

graphs is #P-complete.

  • [Fisher-Temperley 1961, Kasteleyn 1961]: Counting perfect

matchings in planar graphs is in P.

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More generally, if every edge e of G has some weight w(e), define PerfMatch(G) =

  • perfect matchings M

e∈M

w(e)

  • .

Theorem (FKT algorithm)

If G is a planar weighted graph, PerfMatch(G) can be computed in polynomial time.

Idea.

For an arbitrary graph G with adjacency matrix A, the Pfaffian Pf(A) =

  • perfect matchings M

sgn(M)

e∈M

w(e)

  • satisfies Pf(A)2 = det(A). For planar graphs, can flip the signs
  • f some entries of A to make Pf and PerfMatch agree.
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(x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x3 ∨ x4) ∧ (x5 ∨ x6 ∨ x7) ∧ (x4 ∨ x5 ∨ x6)

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(x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x3 ∨ x4) ∧ (x5 ∨ x6 ∨ x7) ∧ (x4 ∨ x5 ∨ x6)

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Imagine: each vertex v on the left propagates signals along its

  • utgoing edges indicating whether v is assigned 1 (green) or 0

(black).

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Each satisfying assignment corresponds to a collection of signals satisfying two constraints: Consistency: If xi is a vertex

  • n the left, the two signals xi

generates must be the same. Satisfaction: If Cj is a vertex

  • n the right, at least one of the

three signals it receives must be 1. 00 1 01 10 11 1 000 001 1 010 1 011 1 100 1 101 1 110 1 111 1

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Goal: encode these bit vectors using the matching properties of graphs

Definition

A matchgate is a weighted graph G with designated subsets of its vertices called external nodes X. We say that it is of arity |X|.

Definition

The standard signature G of matchgate G of arity n is a vector

  • f dimension 2n with entries indexed by bitstrings of length n.

For Z ⊂ X corresponding to bitstring α, Γα = PerfMatch(Γ\Z).

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00 3 01 10 11 5

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000 001 3 010 3 011 100 3 101 110 111 5

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We want planar matchgates G and R whose standard signatures respectively match the vectors encoding the consistency and satisfaction constraints: Consistency: If xi is a vertex

  • n the left, the two signals xi

generates must be the same. Satisfaction: If Cj is a vertex

  • n the right, at least one of the

three signals it receives must be 1. 00 1 01 10 11 1 000 001 1 010 1 011 1 100 1 101 1 110 1 111 1

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(x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x3 ∨ x4) ∧ (x5 ∨ x6 ∨ x7) ∧ (x4 ∨ x5 ∨ x6)

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(x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x3 ∨ x4) ∧ (x5 ∨ x6 ∨ x7) ∧ (x4 ∨ x5 ∨ x6)

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(x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x3 ∨ x4) ∧ (x5 ∨ x6 ∨ x7) ∧ (x4 ∨ x5 ∨ x6)

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Unfortunately, no recognizer has standard signature (0, 1, 1, 1, 1, 1, 1, 1):

Observation (Parity Condition)

Because a graph with an odd number of vertices has no perfect matchings, given any matchgate G, the indices of the nonzero entries in its standard signature must have the same parity.

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  • The saving grace: rewrite number of perfect matchings of

matchgrid Ω as an inner product and apply a change of basis.

  • Suppose there are w wires in Ω, generators G1, ...Gg, and

R1, ..., Rr recognizers, then PerfMatch(Ω) =

  • z∈{0,1}w,

z=x1◦···◦xr◦ y1◦···◦yg

 

g

  • i=1

Giyi

r

  • j=1

Rjxj   = G, R, where G = ⊗iGi and R = ⊗iRi with the order of tensoring specified by the wires.

  • Regard G as an element in X = C2w and R as an element

in X∗: PerfMatch(Ω) is the result of applying dual vector R to G, which is independent of the choice of basis for X.

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Definition

Given a 2 × 2 basis matrix M, the signature with respect to M

  • f a generator G of arity n is the vector G satisfying

G = M⊗nG. The signature with respect to M of a recognizer R of arity n is the vector R satisfying R = RM⊗n.

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  • Suffices to find a basis M of matchgates G and R whose

signatures with respect to M match the vectors encoding the consistency and satisfaction constraints.

  • Over C and F2, this still cannot be done.
  • [Valiant ’06, Cai-Lu ’07]: Over F7, take M =

1 3 6 5

  • ,

G = (3, 0, 0, 5), and R = (0, 3, 3, 0, 3, 0, 0, 5).

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00 3 01 10 11 5 000 001 3 010 3 011 100 3 101 110 111 5

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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  • The number of different values that objects in a counting

problem can take on is called the domain size.

  • Domain size 2:

◮ Boolean satisfying assignments ◮ Vertex covers ◮ Perfect matchings ◮ Ice problems

  • Domain size k

◮ k-colorings

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Over domain size k:

  • Arity-n signatures are now vectors of dimension kn.
  • M now has width k because

G = M⊗nG R = RM⊗n.

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  • Domain size 2: encode True/False by presence/absence
  • f one external node
  • Domain size k: encode colors {1, ..., k} by removal of some

subset of a group of ℓ external nodes

◮ Arities are now multiples of ℓ ◮ External nodes grouped into blocks of ℓ, with wires

connecting matchgates blockwise.

◮ If Γ has n blocks, Γ has 2ℓn entries. ◮ M has height 2ℓ because

G = M ⊗nG R = RM ⊗n.

◮ We call ℓ the basis size.

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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We will regard standard signatures as matrices:

Definition

For standard signature G of generator G, the t-th matrix form G(t) (1 ≤ t ≤ n) is the 2ℓ × 2(n−1)ℓ matrix of entries of G where the rows are indexed by αt ∈ {0, 1}ℓ and the columns are indexed by α1 · · · αt−1αt+1 · · · αn ∈ {0, 1}(n−1)ℓ.

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We will also regard signatures as matrices:

Definition

For signature G of generator G, the t-th matrix form G(t) (1 ≤ t ≤ n) is the k × kn−1 matrix of entries of G where the rows are indexed by αt ∈ [k] and the columns are indexed by α1 · · · αt−1αt+1 · · · αn ∈ [k]n−1. Note: we will denote row indices by superscripts and column indices by subscripts.

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Definition

A generator G is full rank if there exists t for which rank(G(t)) = k. It turns out we may assume that rank(M) = k. But we know G(t) = MG(t)(MT )⊗(n−1). So if G is of full rank, rank(G(t)) = k.

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Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size:

Question

Given k, what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ? domain size basis size Cai-Lu ’08 2 1

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Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size:

Question

Given k, what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ? domain size basis size Cai-Lu ’08 2 1 Cai-Fu ’14 3 1 4 2

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Key to understanding the ultimate capabilities of holographic algorithms for solving counting problems over a given domain size:

Question

Given k, what is the smallest ℓ for which any holographic algorithm over domain size k with a full-rank matchgate can be simulated by one with basis size ℓ? domain size basis size Cai-Lu ’08 2 1 Cai-Fu ’14 3 1 4 2 C ’15, Xia ’15 k ⌊log2 k⌋

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Definition

Z ⊂ {0, 1}n is a cluster if there exists s ∈ {0, 1}n and positions p1, ..., pm ∈ [n] such that each member of Z is of the form s ⊕

  • j∈J epj
  • for some J ⊂ {p1, ..., pm}, where epj is the

bitstring consisting of zeroes everywhere except position pj. We write Z as s + {ep1, ..., epm} (s only unique up to the bits

  • utside of positions p1, ..., pm).

e.g. {000, 001, 100, 101} is a cluster denoted 000 + {e1, e3}.

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For now, assume k = 2K. Steps of proof:

  • 1. Cluster existence: Any standard signature of rank at

least 2K contains a cluster of 2K linearly independent rows

  • 2. Group property: Inverses of standard signatures are also

standard signatures

  • 3. Simulation: Use 1 and 2 to simulate with a basis of size

K

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For now, assume k = 2K. Steps of proof:

  • 1. Cluster existence: Any standard signature of rank at

least 2K contains a cluster of 2K linearly independent rows (hardest part of the proof )

  • 2. Group property: Inverses of standard signatures are also

standard signatures

  • 3. Simulation: Use 1 and 2 to simulate with a basis of size

K

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For now, assume k = 2K. Steps of proof:

  • 1. Cluster existence: Any standard signature of rank at

least 2K contains a cluster of 2K linearly independent rows (hardest part of the proof )

  • 2. Group property: Inverses of standard signatures are also

standard signatures (adapted from Li/Xia result in matchgate character theory)

  • 3. Simulation: Use 1 and 2 to simulate with a basis of size

K

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For now, assume k = 2K. Steps of proof:

  • 1. Cluster existence: Any standard signature of rank at

least 2K contains a cluster of 2K linearly independent rows (hardest part of the proof )

  • 2. Group property: Inverses of standard signatures are also

standard signatures (adapted from Li/Xia result in matchgate character theory)

  • 3. Simulation: Use 1 and 2 to simulate with a basis of size

K (technique due to Cai/Fu)

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Lemma (Group Property)

Full-rank 2K × 2(n−1)K standard signatures G(t) have right inverses (under matrix multiplication) that are also standard signatures.

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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We use the approach introduced by Cai-Fu for simulation given cluster existence and group property have been proven. Take any holographic algorithm over domain size k = 2K.

  • 1. By cluster existence, can pick out a generator G with

full-rank signature and find a cluster Z = s + {ep1, ..., epK}

  • f 2K linearly independent rows. Suppose WLOG s = 0ℓ.
  • 2. Let MZ denote the submatrix of M with rows indexed by
  • Z. This will be the basis of size log k = K we use for the

simulation.

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G∗←Z = (M Z)⊗nG Gtc←Z = (M Z)⊗(t−1) ⊗M ⊗(M Z)⊗(n−t)G

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  • Modifying generators is easy:

G∗←Z

i

= (MZ)⊗niGi has signature Gi with respect to new basis MZ

  • Modifying recognizers is more subtle. Can write

Rj =

  • Rj(M/MZ)⊗mi
  • (MZ)⊗mj.

Is Rj(M/MZ)⊗mi a valid recognizer standard signature?

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  • 3. Define T = M(MZ)−1.
  • 4. By construction,

Gtc←Z(t) = TG∗←Z(t).

  • 5. By group property, G∗←Z(t) has a right-inverse, so

right-multiply by this on both sides to conclude that T is a standard signature.

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Over the new basis MZ:

  • 6. Replace each recognizer Ri with RiT ⊗mi
  • 7. Replace each generator Gj with G∗←Z

j

.

  • 8. These new matchgates have the same signatures as the
  • riginals, but over a basis of size K, so we’re done.
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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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The key ingredient:

Theorem (Rank Rigidity)

The rank of any standard signature Γ (in matrix form) is always a power of two.

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Our methods are primarily algebraic and rely on the characterization of the set of all standard signatures as the variety cut out by a certain collection of quadratic relations:

Theorem (Matchgate Identities)

A 2ℓ × 2(n−1)ℓ matrix Γ is the t-th matrix form of the standard signature of some generator matchgate iff for all ζ, η ∈ {0, 1}(n−1)ℓ and σ, τ ∈ {0, 1}ℓ, the following matchgate identity (MGI) holds. Let ζ ⊕ η = eq1 ⊕ · · · ⊕ eqd′ and σ ⊕ τ = ep1 ⊕ · · · ⊕ epd, where q1 < · · · < qd′ and p1 < · · · < pd. Then if d is even,

d

  • i=1

(−1)i+1Γ

(σ⊕ep1⊕epi) ζ

Γ

(τ⊕ep1⊕epi) η

= ±

d′

  • j=1

(−1)j+1Γ

(σ⊕ep1) (ζ⊕eqj ) Γ (τ⊕ep1) (η⊕eqj ).

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d

  • i=1

(−1)i+1Γ

(σ⊕ep1⊕epi) ζ

Γ

(τ⊕ep1⊕epi) η

= ±

d′

  • j=1

(−1)j+1Γ

(σ⊕ep1) (ζ⊕eqj ) Γ (τ⊕ep1) (η⊕eqj ).

Definition

A 2ℓ × 2m matrix M is a pseudo-signature if for all σ, τ for which wt(σ ⊕ τ) is even, its entries satisfy the corresponding MGI up to a factor of ±1 on the right-hand side. E.g. (matrix-form) standard signatures, clusters of rows, and their transposes are all pseudo-signatures.

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The matchgate identities allow us to deduce key linear algebraic relationships between the rows of any pseudo-signature Γ.

Example

Suppose that d = 2, σ = 0000, τ = 0011, ζ = 1100, η = 1111. Then the MGIs become Γ0000

1100Γ0011 1111 − Γ0011 1100Γ0000 1111 = ±

  • Γ0001

1101Γ0010 1110 − Γ0001 1110Γ0010 1101

  • .

1100 1101 1110 1111 0000 0001 0010 0011

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Example (Cont’d)

  • Rows Γ1100 and Γ1111 are linearly dependent if Γ1101 and

Γ1110 are linearly dependent.

  • Similarly, rows Γ0000 and Γ1111 are linearly dependent if

◮ Γ0001 and Γ1110 ◮ Γ0010 and Γ1101 ◮ Γ0100 and Γ1011 ◮ Γ1000 and Γ0111

are linearly dependent

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Lemma

Let σ, τ ∈ {0, 1}ℓ be such that σ ⊕ τ = 2d

j=1 epi. If row Γ(σ⊕epi)

is linearly dependent with row Γ(τ⊕epi) for all 1 ≤ i ≤ 2d, then row Γσ is linearly dependent with row Γτ. Coordinate-free interpretation: linear relations among wedges of rows of even parity yield linear relations among wedges of rows

  • f odd parity.
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Definition

For V a vector space with basis {ej}, the second exterior power

  • f V , denoted Λ2V , is the vector space given by quotienting

V ⊗ V by the relation v ⊗ w ∼ −w ⊗ v for all v, w ∈ V . We denote the image of v ⊗ w under this quotient map by v ∧ w. Λ2V has basis {ei ∧ ej}i<j. Explicitly, if v = viei and w = wiei, then v ∧ w =

  • i<j

(viwj − vjwi)ei ∧ ej =

  • i<j
  • vi

vj wi wj

  • ei ∧ ej.

In particular, v and w are linearly dependent iff v ∧ w = 0.

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By the MGIs, linear relations among wedges of rows of even parity yield linear relations among wedges of rows of odd parity.

Example

Γ0000 ∧ Γ1111 = 0 implies that Γ0001 ∧ Γ1110 − Γ0010 ∧ Γ1101 + Γ0100 ∧ Γ1011 − Γ1000 ∧ Γ0111 = 0

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By the MGIs, linear relations among wedges of rows of even parity yield linear relations among wedges of rows of odd parity.

Example

µ · (Γ0000 ∧ Γ1111) + ν · (Γ0011 ∧ Γ1100) = 0 implies that µ · (Γ0001 ∧ Γ1110 − Γ0010 ∧ Γ1101 + Γ0100 ∧ Γ1011 − Γ1000 ∧ Γ0111)± ν ·(Γ0010 ∧Γ1101 −Γ0001 ∧Γ1110 +Γ0111 ∧Γ1000 −Γ1011 ∧Γ0100) = 0

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Claim

Rank rigidity implies cluster existence.

Proof.

Suppose Γ is a 2ℓ × 2m pseudo-signature of rank 2K. Can assume Γ has no proper clusters of rows with the same rank as Γ. Otherwise, if there were such a cluster Z = σ + {eq1, ..., eqℓ′}, replace Γ by ΓZ and ℓ by ℓ′, and ignore bits in positions outside of q1, ..., qℓ′.

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111

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ℓ−K+1=1

  • K−1
  • 0000

0001 . . . . . . 1111

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1110 0101

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1110 0101

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000 1111 0001 . . . . . . 1111 1111

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000 1111 0001 . . . . . . 1111 1111 1110 1010

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000 1111 0001 . . . . . . 1111 1111 1110 1010

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000 1111 0001 . . . . . . 1111 1111 1110 1010

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ℓ−K+1

  • 0000

K−1

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000 1111 0001 . . . . . . 1111 1111 All other rows must be zero. By the MGIs, Γ0z◦0K−1 ∧ Γ1z◦0K−1 = 0, a contradiction.

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Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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Theorem

If Γ is a 2K+1 × 2m pseudo-signature with rank at least 2K + 1, then rank(Γ) = 2K+1.

Sketch.

Inductively, we know that Γ contains a cluster Z of 2K linearly independent rows, say 0K+1 ⊕ {e2, ..., eK+1}. Because rank(Γ) ≥ 2K + 1, there exists a row outside the linear span of Z.

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1001 lay in the span of the red rows, so Γ1001 ∧ Γ0000 =

  • σ red, even

aσ ·

  • Γσ ∧ Γ0000

. LHS: Γ1000 ∧ Γ0001 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1010 lay in the span of the red rows, so Γ1010 ∧ Γ0000 =

  • σ red, even

aσ ·

  • Γσ ∧ Γ0000

. LHS: Γ1000 ∧ Γ0010 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ1000 ∧ Γ0001

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1100 lay in the span of the red rows, so Γ1100 ∧ Γ0000 =

  • σ red, even

aσ ·

  • Γσ ∧ Γ0000

. LHS: Γ1000 ∧ Γ0100 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ1000 ∧ Γ0001, Γ1000 ∧ Γ0010

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1011 lay in the span of the red rows, so Γ1011 ∧ Γ0001 =

  • σ red, odd

aσ ·

  • Γσ ∧ Γ0001

. LHS: Γ1001 ∧ Γ0011 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ0000 ∧ Γ1001

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1101 lay in the span of the red rows, so Γ1011 ∧ Γ0001 =

  • σ red, odd

aσ ·

  • Γσ ∧ Γ0001

. LHS: Γ1001 ∧ Γ0101 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ0000 ∧ Γ1001, Γ1001 ∧ Γ0011

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1110 lay in the span of the red rows, so Γ1110 ∧ Γ0001 =

  • σ red, odd

aσ ·

  • Γσ ∧ Γ0001

. LHS: Γ1111 ∧ Γ0000, Γ1100 ∧ Γ0011, Γ1010 ∧ Γ0101, Γ0110 ∧ Γ1001 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ0000 ∧ Γ1001, Γ1001 ∧ Γ0011, Γ1001 ∧ Γ0101

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Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110 Suppose Γ1111 lay in the span of the red rows, so Γ1111 ∧ Γ0000 =

  • σ red, even

aσ ·

  • Γσ ∧ Γ0000

. LHS: Γ1110 ∧ Γ0001, Γ1101 ∧ Γ0010, Γ1011 ∧ Γ0100, Γ0111 ∧ Γ1000 RHS: Γ0∗∗∗ ∧ Γ0∗∗∗, Γ1000 ∧ Γ0001, Γ1000 ∧ Γ0010, Γ1000 ∧ Γ0100

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SLIDE 82

Even columns: 0000 0011 0101 0110 1001 1010 1100 1111 Odd columns: 0001 0010 0100 0111 1000 1011 1101 1110

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SLIDE 83

Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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SLIDE 84

Theorem

Suppose ℓ > K + 1. If Γ is a 2ℓ × 2m pseudo-signature of rank ≥ 2K + 1, then there exists a cluster Z {0, 1}ℓ for which ΓZ is also of rank ≥ 2K + 1. Suppose to the contrary. Inductively we know Γ has a cluster Z

  • f 2K linearly independent rows, say 0ℓ + {ep1, ..., epK}.
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SLIDE 85

ℓ−K

  • 0000

K

  • 0000

0000 0001 . . . . . . 0000 1111

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SLIDE 86

ℓ−K

  • 0000

K

  • 0000

0000 0001 . . . . . . 0000 1111 1110 0101

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SLIDE 87

ℓ−K

  • 0000

K

  • 0000

0000 0001 . . . . . . 0000 1111 1110 0101

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SLIDE 88

ℓ−K

  • 0000

K

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000

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SLIDE 89

ℓ−K

  • 0000

K

  • 0000

0000 0001 . . . . . . 0000 1111 1111 0000

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SLIDE 90

To show Γ0ℓ and Γ1ℓ−K◦0ℓ are linearly dependent, by MGIs it’s enough to show:

Lemma

Γ0ℓ⊕ej = 0 for all j = p1, ..., pK.

Proof.

For i ∈ {p1, ..., pK} and j ∈ {p1, ..., pK}, define:

  • Ti: all rows u for which ui = 0
  • T j

i : all rows u for which ui = uj = 0

  • Zi: Z ∩ Ti

Note that Zi ⊂ T j

i ⊂ Ti.

So inductively, proper cluster T j

i has rank a power of two, either

2K−1 or 2K.

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SLIDE 91

Proof (Cont’d).

  • If rank(T j

i ) = 2K−1, then span(T j i ) = span(Zi). This is true

for all i ∈ {p1, ..., pK}, so Γ0ℓ⊕ej ∈ ∩K

i=1span(Zi) = span({Γ0ℓ}).

  • If rank(T j

i ) = 2K, then span(Ti) = span(T j i ) ⊂ span(Z), a

contradiction.

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SLIDE 92

Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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SLIDE 93

Theorem (Fu/Yang ’13)

Suppose a basis collapse theorem holds on domain size 2. Then if a holographic algorithm uses a 2ℓ × k basis of rank 2, then the same collapse theorem holds for this holographic algorithm.

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SLIDE 94

Theorem

Suppose a basis collapse theorem holds on domain size r. Then if a holographic algorithm uses a 2ℓ × k basis of rank r, then the same collapse theorem holds for this holographic algorithm.

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SLIDE 95

By rank rigidity, G(t) must have rank a power of two. If G is a full-rank signature, by G(t) = MG(t)(MT )⊗(n−1), we know M must have rank a power of two. So if k = 2K, we’re done inductively by the collapse theorem for domain size 2K, where K = ⌊log2 k⌋.

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SLIDE 96

Introduction Matchgates/Holographic Algorithms: A Crash Course Basis Size and Domain Size Collapse Theorems Setup Overview of Proof Group Property Simulation Rank Rigidity Matchgate Identities Implies Cluster Existence Base Case Inductive Step Epilogue k = 2K? Next Steps Acknowledgments

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SLIDE 97
  • Work out the case where no full-rank matchgate exists
  • Use the collapse theorem to initiate a study of holographic

algorithms over higher domains

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SLIDE 98

Thank you to:

  • Professor Leslie Valiant (Harvard University)
  • Professor Jin-Yi Cai (University of Wisconsin)
  • Harvard Herchel-Smith Research Fellowship
  • Simons Institute for Computing